如何迭代由空格分隔的单词组成的字符串中的单词?

注意,我对C字符串函数或那种字符操作/访问不感兴趣。比起效率,我更喜欢优雅。我当前的解决方案:

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

int main() {
    string s = "Somewhere down the road";
    istringstream iss(s);

    do {
        string subs;
        iss >> subs;
        cout << "Substring: " << subs << endl;
    } while (iss);
}

当前回答

我使用以下代码:

namespace Core
{
    typedef std::wstring String;

    void SplitString(const Core::String& input, const Core::String& splitter, std::list<Core::String>& output)
    {
        if (splitter.empty())
        {
            throw std::invalid_argument(); // for example
        }

        std::list<Core::String> lines;

        Core::String::size_type offset = 0;

        for (;;)
        {
            Core::String::size_type splitterPos = input.find(splitter, offset);

            if (splitterPos != Core::String::npos)
            {
                lines.push_back(input.substr(offset, splitterPos - offset));
                offset = splitterPos + splitter.size();
            }
            else
            {
                lines.push_back(input.substr(offset));
                break;
            }
        }

        lines.swap(output);
    }
}

// gtest:

class SplitStringTest: public testing::Test
{
};

TEST_F(SplitStringTest, EmptyStringAndSplitter)
{
    std::list<Core::String> result;
    ASSERT_ANY_THROW(Core::SplitString(Core::String(), Core::String(), result));
}

TEST_F(SplitStringTest, NonEmptyStringAndEmptySplitter)
{
    std::list<Core::String> result;
    ASSERT_ANY_THROW(Core::SplitString(L"xy", Core::String(), result));
}

TEST_F(SplitStringTest, EmptyStringAndNonEmptySplitter)
{
    std::list<Core::String> result;
    Core::SplitString(Core::String(), Core::String(L","), result);
    ASSERT_EQ(1, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
}

TEST_F(SplitStringTest, OneCharSplitter)
{
    std::list<Core::String> result;

    Core::SplitString(L"x,y", L",", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L"y", *result.rbegin());

    Core::SplitString(L",xy", L",", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
    ASSERT_EQ(L"xy", *result.rbegin());

    Core::SplitString(L"xy,", L",", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"xy", *result.begin());
    ASSERT_EQ(Core::String(), *result.rbegin());
}

TEST_F(SplitStringTest, TwoCharsSplitter)
{
    std::list<Core::String> result;

    Core::SplitString(L"x,.y,z", L",.", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L"y,z", *result.rbegin());

    Core::SplitString(L"x,,y,z", L",,", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L"y,z", *result.rbegin());
}

TEST_F(SplitStringTest, RecursiveSplitter)
{
    std::list<Core::String> result;

    Core::SplitString(L",,,", L",,", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
    ASSERT_EQ(L",", *result.rbegin());

    Core::SplitString(L",.,.,", L",.,", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
    ASSERT_EQ(L".,", *result.rbegin());

    Core::SplitString(L"x,.,.,y", L",.,", result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L".,y", *result.rbegin());

    Core::SplitString(L",.,,.,", L",.,", result);
    ASSERT_EQ(3, result.size());
    ASSERT_EQ(Core::String(), *result.begin());
    ASSERT_EQ(Core::String(), *(++result.begin()));
    ASSERT_EQ(Core::String(), *result.rbegin());
}

TEST_F(SplitStringTest, NullTerminators)
{
    std::list<Core::String> result;

    Core::SplitString(L"xy", Core::String(L"\0", 1), result);
    ASSERT_EQ(1, result.size());
    ASSERT_EQ(L"xy", *result.begin());

    Core::SplitString(Core::String(L"x\0y", 3), Core::String(L"\0", 1), result);
    ASSERT_EQ(2, result.size());
    ASSERT_EQ(L"x", *result.begin());
    ASSERT_EQ(L"y", *result.rbegin());
}

其他回答

这里有一个拆分函数:

是通用的使用标准C++(无增强)接受多个分隔符忽略空标记(可以轻松更改)模板<typename T>矢量<T>拆分(常量T&str,常量T&分隔符){向量<T>v;typename T::size_type start=0;自动位置=str.find_first_of(分隔符,开始);而(pos!=T::npos){if(pos!=开始)//忽略空标记v.template_back(str,start,pos-start);开始=位置+1;pos=str.find_first_of(分隔符,开始);}if(start<str.length())//忽略尾随分隔符v.template_back(str,start,str.length()-start);//添加字符串的剩余部分返回v;}

示例用法:

    vector<string> v = split<string>("Hello, there; World", ";,");
    vector<wstring> v = split<wstring>(L"Hello, there; World", L";,");
#include <iostream>
#include <regex>

using namespace std;

int main() {
   string s = "foo bar  baz";
   regex e("\\s+");
   regex_token_iterator<string::iterator> i(s.begin(), s.end(), e, -1);
   regex_token_iterator<string::iterator> end;
   while (i != end)
      cout << " [" << *i++ << "]";
}

IMO,这是最接近python的re.split()。有关regex_token_iterator的更多信息,请参阅cplusplus.com。-1(regex_token_iterator tor中的第四个参数)是序列中不匹配的部分,使用匹配作为分隔符。

我用这个分隔符分隔字符串。第一个将结果放入预先构建的向量中,第二个返回新向量。

#include <string>
#include <sstream>
#include <vector>
#include <iterator>

template <typename Out>
void split(const std::string &s, char delim, Out result) {
    std::istringstream iss(s);
    std::string item;
    while (std::getline(iss, item, delim)) {
        *result++ = item;
    }
}

std::vector<std::string> split(const std::string &s, char delim) {
    std::vector<std::string> elems;
    split(s, delim, std::back_inserter(elems));
    return elems;
}

请注意,此解决方案不会跳过空令牌,因此下面将找到4项,其中一项为空:

std::vector<std::string> x = split("one:two::three", ':');

这里有一个仅使用标准正则表达式库的正则表达式解决方案。(我有点生疏,所以可能会有一些语法错误,但这至少是一般的想法)

#include <regex.h>
#include <string.h>
#include <vector.h>

using namespace std;

vector<string> split(string s){
    regex r ("\\w+"); //regex matches whole words, (greedy, so no fragment words)
    regex_iterator<string::iterator> rit ( s.begin(), s.end(), r );
    regex_iterator<string::iterator> rend; //iterators to iterate thru words
    vector<string> result<regex_iterator>(rit, rend);
    return result;  //iterates through the matches to fill the vector
}

这是我解决这个问题的方法:

vector<string> get_tokens(string str) {
    vector<string> dt;
    stringstream ss;
    string tmp; 
    ss << str;
    for (size_t i; !ss.eof(); ++i) {
        ss >> tmp;
        dt.push_back(tmp);
    }
    return dt;
}

此函数返回字符串向量。